Worked solution · Exam P

SOA Exam P Sample Question #377 Solution

This Exam P sample reference tests Conditional Probability. This is a conditional probability within the subgroup of policyholders receiving the specified treatment. Of the 88 treated members, 28 belong to the low-or-normal-rate category, so the requested probability is 28/88=7/22, approximately 0.318, and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Conditional Probability question

Setup

Conditioning on treatment restricts the denominator to treated policyholders from both heart-rate categories.

n_{\mathrm{treated}}=60+28=88

Model

Within that restricted group, the favorable count is the number from the low-or-normal category.

\Pr(L\mid S)=\frac{n(L\cap S)}{n(S)}

\Pr(L\mid S)=\frac{28}{60+28}

Compute

Reduce the count ratio and convert it to a decimal.

\frac{28}{88}=\frac7{22}=0.3181818182

Answer

Approximately 31.8% of the treated subgroup belongs to the requested category.

\boxed{0.318\quad\text{(C)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThe ratio 28/3000=0.00933 is the unconditional joint probability of being in the low-or-normal category and receiving treatment. It does not condition on treatment.
BThe ratio 28/1900, or 0.01474, is the treatment probability within the low-or-normal group. This reverses the requested conditioning.
DThe ratio 28/60=0.46667 compares the two favorable cell counts but does not divide by the total treated count 28+60.
EThe ratio 1900/3000=0.63333 is the unconditional proportion in the low-or-normal category; it ignores the treatment information.

Original practice · fully worked

Original variant: extreme readings after a high alert

Three independent sensor readings are uniformly distributed from 0 to 1. The system reports that the largest reading exceeded 0.80. Given this report, calculate the probability that the smallest reading was below 0.20.

A
0.192
B
0.216
C
0.393
D
0.488
E
0.607

Answer and calculation

All three readings stay at or below 0.80 with probability 0.512, so the high-alert probability is 0.488. Inclusion-exclusion gives joint probability 0.192 for a reading below 0.20 and another above 0.80. Dividing by the alert probability gives 24/61, so choice C.

Setup

Let H be the event that the maximum exceeds 0.80 and L the event that the minimum is below 0.20.

\Pr(H)=1-(0.80)^3=0.488

\Pr(L)=1-(0.80)^3=0.488

Model

Use inclusion-exclusion. The complement of H keeps every reading at most 0.80, while the complement of L keeps every reading at least 0.20.

\Pr(H\cap L)=1-\Pr(H^c)-\Pr(L^c)+\Pr(H^c\cap L^c)

\Pr(H^c\cap L^c)=(0.80-0.20)^3=0.216

Compute

Calculate the joint event and divide by the reporting probability.

\Pr(H\cap L)=1-0.512-0.512+0.216=0.192

\Pr(L\mid H)=\frac{0.192}{0.488}=\frac{24}{61}=0.3934426230

Answer

The conditional probability is approximately 0.393.

\boxed{0.393\quad\text{(C)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: extreme readings after a high alert

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability